MCQ The nth derivative of $f(x)=\frac{1+x}{1-x}$

Question

Let $f(x)=\dfrac{1+x}{1-x}$ The nth derivative of f is equal to:

$\dfrac{2n}{(1-x)^{n+1}} $
$\dfrac{2(n!)}{(1-x)^{2n}} $
$\dfrac{2(n!)}{(1-x)^{n+1}} $

by Leibniz formula

$$ {\displaystyle \left( \dfrac{1+x}{1-x}\right)^{(n)}=\sum _{k=0}^{n}{\binom {n}{k}}\ (1+x)^{(k)}\ \left(\dfrac{1}{1-x}\right)^{(n-k)}}$$

using the hint

$\dfrac{1+x}{1-x}=\dfrac{2-(1-x)}{1-x}=\dfrac2{1-x}-1$ and
$\left(\dfrac{1}{x}\right)^{n}=\dfrac{(-1)^{n}n!}{x^{n+1}}$

so

$${\displaystyle \left( \dfrac{1+x}{1-x} \right)^{(n)} = \left( \dfrac{2}{1-x}-1 \right)^{(n)}=2\dfrac{ (-1)^{n}n! }{ (1-x)^{n+1} } } $$ but this result isn't apear in any proposed answers

what about the method of Lord Shark the Unknown

tell me please this way holds for any mqc question contain find the n th derivative so it's suffice to check each answer in y case i will start with first

let $f_n(x)=\dfrac{2n}{(1-x)^{n+1}}$ then $f_{n+1}(x)=\dfrac{2(n+1)}{(1-x)^{n+2}}$ do i have $f'_{n}=f_{n+1}$ let calculate $$ f'_n=\dfrac{-2n(n+1)}{(1-x)^{n+2}}\neq f_{n+1}$$
let $f_n(x)=\dfrac{2(n!)}{(1-x)^{2n}}$ then $f_{n+1}(x)=\dfrac{2((n+1)!)}{(1-x)^{2(n+1)}}$ do i have $f'_{n}=f_{n+1}$ let calculate $$ f'_n=\dfrac{-2(n!)(2n)}{(1-x)^{4n}}\neq f_{n+1}$$
let $f_n(x)=\dfrac{2(n!)}{(1-x)^{n+1}}$ then $f_{n+1}(x)=\dfrac{2((n+1)!)}{(1-x)^{n+2}}$ do i have $f'_{n}=f_{n+1}$ let calculate $$ f'_n=\dfrac{2(n!)(n+1)}{((1-x)^{n+1})^{2}}=\dfrac{2((n+1)!)}{(1-x)^{2n+2}}\neq f_{n+1}$$

How did you narrow it down to just one of those three possibilities? — Angina Seng, Jul 10 '17 at 06:45
In the last step you made a mistake, you have to apply the chain rule correctly. — Levi Haunschmid, Jul 10 '17 at 07:04

score 5 · Answer 1 · answered Jul 10 '17 at 06:42

5

$$y=\dfrac{1+x}{1-x}=\dfrac{2-(1-x)}{1-x}=\dfrac2{1-x}-1$$

$$\implies\dfrac{dy}{dx}=\dfrac{2(-1)}{(1-x)^2}$$

$$\dfrac{d^2y}{dx^2}=\dfrac{2(-1)(-2)}{(1-x)^3}=\dfrac{2(-1)^22!}{(1-x)^{2+1}}$$

Can you follow the pattern?

answered Jul 10 '17 at 06:42

lab bhattacharjee

274,582

check my update – Cyro Jul 10 '17 at 06:53

score 3 · Answer 2 · answered Jul 10 '17 at 06:47

3

If we let $f_n$ denote the $n$-th derivative, then $f_{n+1}=f_n'$. That is the case only for one of the three possible solutions you have there.

answered Jul 10 '17 at 06:47

Angina Seng

158,341

let $f_n(x)=\dfrac{2n}{(1-x)^{n+1}}$ then $f_{n+1}(x)=\dfrac{2(n+1)}{(1-x)^{n+2}}$ do i have $f'{n}=f{n+1}$ let calculate $$ f'n=\dfrac{-2n(n+1)}{(1-x)^{n+2}}\neq f{n+1}$$
let $f_n(x)=\dfrac{2(n!)}{(1-x)^{2n}}$ then $f_{n+1}(x)=\dfrac{2((n+1)!)}{(1-x)^{2(n+1)}}$ do i have $f'{n}=f{n+1}$ let calculate $$ f'n=\dfrac{-2(n!)(2n)}{(1-x)^{4n}}\neq f{n+1}$$
let $f_n(x)=\dfrac{2(n!)}{(1-x)^{n+1}}$ then $f_{n+1}(x)=\dfrac{2((n+1)!)}{(1-x)^{n+2}}$ do i have $f'{n}=f{n+1}$ let calculate $$ f'n=\dfrac{2(n!)(n+1)}{((1-x)^{n+1})^{2}}=\dfrac{2((n+1)!)}{(1-x)^{2n+2}}\neq f{n+1}$$

Cyro

Jul 10 '17 at 07:22

@Cyro The first and third examples differ only by a scalar multiple, yet your derivatives do not. – Angina Seng Jul 10 '17 at 17:37

score 0 · Answer 3 · answered Jul 10 '17 at 06:55

0

You can do it with Leibniz's formula (not that it's easier than without it); just consider that $$ (1+x)^{(k)}= \begin{cases} 1+x & k=0 \\ 1 & k=1 \\ 0 & k>1 \end{cases} $$ so the formula gives $$ \left( \frac{1+x}{1-x}\right)^{\!(n)}= (1+x)\left(\dfrac{1}{1-x}\right)^{\!(n)}+ n\left(\frac{1}{1-x}\right)^{\!(n-1)} $$ Now make a conjecture about $$ \left(\frac{1}{1-x}\right)^{\!(n)} $$

answered Jul 10 '17 at 06:55

egreg

238,574

$$\left(\dfrac{1}{1-x}\right)^{n}=\dfrac{(-1)^{n}n!}{(1-x)^{n+1}}$$ – Cyro Jul 10 '17 at 07:00
That derivative is incorrect, you are missing another $(-1)^n$ – Levi Haunschmid Jul 10 '17 at 07:05
@Cyro Try the first and second derivatives. You'll see that there is no minus sign in front. – egreg Jul 10 '17 at 07:22

MCQ The nth derivative of $f(x)=\frac{1+x}{1-x}$

3 Answers3